How do you want me to give you points tell me and I’ll do it
To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.
The vector product between the Force and the radius allows us to obtain the torque, in this way,





Therefore the torque on the particle about the origen is 50k
PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,





Therefore the angle between the ratio and the force is 103.88°
Edwin Hubble calculated the expansion rate of the
universe. The evidence that he base his calculation is the differences in
redshift for galaxies. The answer is letter B. the red shift of galaxies was
directly proportional to the distance of the galaxy from earth. It means that bodies farther away from Earth
were moving away faster. The Hubble’s constant is the ratio of distance to
redshift equal to 170 kilometers per second per light year of distance.
Answer:
(a) Yes, it is possible by raising the object to a greater height without acceleration.
Explanation:
The work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in kinetic energy requires a change in velocity.
If kinetic energy will not change, then velocity will not change, this means that there will be constant velocity and an object with a constant velocity is not accelerating.
If the object is not accelerating (without acceleration) and it remains at the same height (change in height = 0, and mgh = 0).
Thus, for work to be done on the object, without changing the kinetic energy of the object, the object must be raised to a greater height without acceleration.
Correct option is " (a) Yes, it is possible by raising the object to a greater height without acceleration".
The water and solutes the first enters the nephron tubules undergo a process just before it exits the body as a urine. This process is called filtration. It operates when 20 percent of the solution volume is filtered in the renal corpuscle at a given time. It is reportedly stated that there are around 180 liters of this fluid being filtered by the kidney every day.